The present invention relates to a semiconductor design technology; and more particularly, to a technology of removing crosstalk that may occur between transmission lines of the semiconductor device.
In semiconductor devices, when a signal is transmitted via a transmission line, the signal may be affected by electromagnetic fields, which will be discussed in detail below.
When a dielectric is between metals, there is capacitance between them. In general, a transmission line is made of metal and a dielectric is arranged between transmission lines, and thus, capacitance may exist between them. It is known that even air is a dielectric whose relative dielectric constant is ‘1’. Therefore, even when transmission lines are arranged in the air, there may also be capacitance between them. When an alternate current of high frequency band flows onto a transmission line, electrical energy interference occurs due to the influence of capacitance between transmission lines, i.e., mutual capacitance, as the current goes toward a higher frequency band. This also affects the characteristic impedance value of transmission line, and so on. In addition, as an alternate current flows onto a transmission line, a magnetic field is induced on the transmission line, which produces mutual inductance that affects a magnetic field of another transmission line. This mutual inductance causes magnetic energy interference that affects the inductance value of each transmission line, etc, which in turn affects the characteristic impedance value of each transmission line.
Meanwhile, coupling is a phenomenon in which alternate energies such as an electric field and a magnetic field are mutually transferred between separated spaces or transmission lines. As metals are closed to each other, this phenomenon gives rise to interference with a signal and acts as an undesired parasitic effect. This unnecessary coupling is often called “crosstalk” in view of ElectricMagnetic Interference (EMI). In the invention, therefore, unnecessary mutual interference caused by coupling is defined as crosstalk, which will be set forth below.
As noted earlier, electromagnetic interference that occurs between adjacent transmission lines is referred to as “crosstalk”. This phenomenon arises due to both mutual inductance and mutual capacitance. Such mutual inductance and mutual capacitance have influence on the total inductance and the total capacitance of transmission line. The following is an explanation about the influence of mutual inductance and mutual capacitance in each mode, in an ODD mode and EVEN mode analyze in a coupled transmission line theory.
FIG. 1 shows an equivalent circuit model of a coupled transmission line, which includes an equivalent circuit 100 of inductance and capacitance, an inductance equivalent circuit 110 and a capacitance equivalent circuit 120.
Signal modes between adjacent transmission lines can be largely classified into an ODD mode and an EVEN mode. When there are two transmission lines, the ODD mode is when signals having a 180-degree phase difference, but the same amplitude, are applied to the two transmission lines, respectively. First, regarding inductance, a voltage is generated by inductive coupling and currents I1 and I2 flowing onto two transmission lines in the inductance equivalent circuit 110 have the same amplitude but opposite directions to each other. Assuming that the self-inductance L11=L22=L0 and the mutual inductance L12=LM, V1 and V2 of the inductance equivalent circuit 110 may be expressed as:
                              V          1                =                                            L              0                        ⁢                                          ⅆ                                  I                  1                                                            ⅆ                t                                              +                                    L              M                        ⁢                                          ⅆ                                  I                  2                                                            ⅆ                t                                                                        Eq        .                                  ⁢                  (          1          )                                                  V          2                =                                            L              0                        ⁢                                          ⅆ                                  I                  2                                                            ⅆ                t                                              +                                    L              M                        ⁢                                          ⅆ                                  I                  1                                                            ⅆ                t                                                                        Eq        .                                  ⁢                  (          2          )                    
In the ODD mode, I1=−I2 and V1=−V2, and thus, these equations may be represented as follows:
                              V          1                =                                                            L                0                            ⁢                                                ⅆ                                      I                    1                                                                    ⅆ                  t                                                      +                                          L                M                            ⁢                                                ⅆ                                      (                                          -                                              I                        1                                                              )                                                                    ⅆ                  t                                                              =                                    (                                                L                  0                                -                                  L                  M                                            )                        ⁢                                          ⅆ                                  I                  1                                                            ⅆ                t                                                                        Eq        .                                  ⁢                  (          3          )                                                  V          2                =                                                            L                0                            ⁢                                                ⅆ                                      I                    2                                                                    ⅆ                  t                                                      +                                          L                M                            ⁢                                                ⅆ                                      (                                          -                                              I                        2                                                              )                                                                    ⅆ                  t                                                              =                                    (                                                L                  0                                -                                  L                  M                                            )                        ⁢                                          ⅆ                                  I                  2                                                            ⅆ                t                                                                        Eq        .                                  ⁢                  (          4          )                                                  L          ODD                =                              L            11                    -                      L            M                                              Eq        .                                  ⁢                  (          5          )                    
As can be seen from Eq. (5) above, the total inductance LODD in the ODD mode becomes smaller than the self-inductance L11 by the mutual inductance LM.
Similarly, relating to capacitance, assuming that in the capacitance equivalent circuit 120 the self-capacitance C1G=C2G=C0 and the mutual capacitance C12=CM, I1 and I2 of the capacitance equivalent circuit 120 may be expressed as:
                              I          1                =                                                            C                0                            ⁢                                                ⅆ                                      V                    1                                                                    ⅆ                  t                                                      +                                          C                M                            ⁢                                                ⅆ                                      (                                                                  V                        1                                            -                                              V                        2                                                              )                                                                    ⅆ                  t                                                              =                                                    (                                                      C                    0                                    -                                      C                    M                                                  )                            ⁢                                                ⅆ                                      V                    1                                                                    ⅆ                  t                                                      -                                          C                M                            ⁢                                                ⅆ                                      V                    2                                                                    ⅆ                  t                                                                                        Eq        .                                  ⁢                  (          6          )                                                  I          2                =                                                            C                0                            ⁢                                                ⅆ                                      V                    2                                                                    ⅆ                  t                                                      +                                          C                M                            ⁢                                                ⅆ                                      (                                                                  V                        2                                            -                                              V                        1                                                              )                                                                    ⅆ                  t                                                              =                                                    (                                                      C                    0                                    -                                      C                    M                                                  )                            ⁢                                                ⅆ                                      V                    2                                                                    ⅆ                  t                                                      -                                          C                M                            ⁢                                                ⅆ                                      V                    1                                                                    ⅆ                  t                                                                                        Eq        .                                  ⁢                  (          7          )                    
In the ODD mode, I1=−I2 and V1=−V2, and thus, these equations may be defined as:
                              I          1                =                                                            C                0                            ⁢                                                ⅆ                                      V                    1                                                                    ⅆ                  t                                                      +                                          C                M                            ⁢                                                ⅆ                                      (                                                                  V                        1                                            -                                              (                                                  -                                                      V                            1                                                                          )                                                              )                                                                    ⅆ                  t                                                              =                                    (                                                C                  0                                +                                  2                  ⁢                                      C                    M                                                              )                        ⁢                                          ⅆ                                  V                  1                                                            ⅆ                t                                                                        Eq        .                                  ⁢                  (          8          )                                                  I          2                =                                                            C                0                            ⁢                                                ⅆ                                      V                    2                                                                    ⅆ                  t                                                      +                                          C                M                            ⁢                                                ⅆ                                      (                                                                  V                        2                                            -                                              (                                                  -                                                      V                            2                                                                          )                                                              )                                                                    ⅆ                  t                                                              =                                    (                                                C                  0                                +                                  2                  ⁢                                      C                    M                                                              )                        ⁢                                          ⅆ                                  V                  2                                                            ⅆ                t                                                                        Eq        .                                  ⁢                  (          9          )                                                                                            C                ODD                            =                                                                    C                                          1                      ⁢                      G                                                        +                                      2                    ⁢                                          C                      M                                                                      =                                                      C                    11                                    +                                      C                    M                                                                                                                                          C                11                            =                                                C                                      1                    ⁢                    G                                                  +                                  C                  M                                                                                        Eq        .                                  ⁢                  (          10          )                    
As can be seen from Eq. (10) above, the total capacitance CODD in the ODD mode becomes greater than the self-capacitance C1G by 2CM.
Based on the total inductance LODD and the total capacitance CODD in Eqs. (5) and (10) above, ZODD and TDODD may be defined as:
                              Z          ODD                =                                                            L                ODD                                            C                ODD                                              =                                                                      L                  11                                -                                  L                  12                                                                              C                  11                                +                                  C                  12                                                                                        Eq        .                                  ⁢                  (          11          )                                                  TD          ODD                =                                                            L                ODD                            ⁢                              C                ODD                                              =                                                    (                                                      L                    11                                    -                                      L                    12                                                  )                            ⁢                              (                                                      C                    11                                    +                                      C                    12                                                  )                                                                        Eq        .                                  ⁢                  (          12          )                    
Meanwhile, the EVEN mode is when signals having the same phase and the same amplitude are applied to the two transmission lines, respectively. First, as for inductance, a voltage is generated by an inductive coupling and currents I1 and I2 flowing onto the two transmission lines in the inductance equivalent circuit 110 have the same amplitude and the same direction. Assuming that the self-inductance L11=L22=L0 and the mutual inductance L12=LM, V1 and V2 of the inductance equivalent circuit 110 may be expressed as in Eqs. (1) and (2) above. And, in the EVEN mode I1=I2 and V1=V2, and thus, these equations may be represented as:
                              V          1                =                                                            L                0                            ⁢                                                ⅆ                                      I                    1                                                                    ⅆ                  t                                                      +                                          L                M                            ⁢                                                ⅆ                                      (                                          I                      1                                        )                                                                    ⅆ                  t                                                              =                                    (                                                L                  0                                +                                  L                  M                                            )                        ⁢                                          ⅆ                                  I                  1                                                            ⅆ                t                                                                        Eq        .                                  ⁢                  (          13          )                                                  V          2                =                                                            L                0                            ⁢                                                ⅆ                                      I                    2                                                                    ⅆ                  t                                                      +                                          L                M                            ⁢                                                ⅆ                                      (                                          I                      2                                        )                                                                    ⅆ                  t                                                              =                                    (                                                L                  0                                +                                  L                  M                                            )                        ⁢                                          ⅆ                                  I                  2                                                            ⅆ                t                                                                        Eq        .                                  ⁢                  (          14          )                                                  L          EVEN                =                              L            11                    +                      L            M                                              Eq        .                                  ⁢                  (          15          )                    
As can be seen from Eq. (15), the total inductance LEVEN in the EVEN mode becomes the self-inductance L11 plus the mutual inductance LM.
Similarly, assuming that in the capacitance equivalent circuit 120 the capacitance may be expressed as Eqs. (6) and (7) above, and in the EVEN mode I1=I2 and V1=V2, and thus, I1 and I2 may be represented again as:
                              I          1                =                                                            C                0                            ⁢                                                ⅆ                                      V                    1                                                                    ⅆ                  t                                                      +                                          C                M                            ⁢                                                ⅆ                                      (                                                                  V                        1                                            -                                              V                        1                                                              )                                                                    ⅆ                  t                                                              =                                    C              0                        ⁢                                          ⅆ                                  V                  1                                                            ⅆ                t                                                                        Eq        .                                  ⁢                  (          16          )                                                  I          2                =                                                            C                0                            ⁢                                                ⅆ                                      V                    2                                                                    ⅆ                  t                                                      +                                          C                M                            ⁢                                                ⅆ                                      (                                                                  V                        2                                            -                                              V                        2                                                              )                                                                    ⅆ                  t                                                              =                                    C              0                        ⁢                                          ⅆ                                  V                  2                                                            ⅆ                t                                                                        Eq        .                                  ⁢                  (          17          )                                                                                            C                EVEN                            =                                                C                                      1                    ⁢                    G                                                  =                                                      C                    11                                    -                                      C                    M                                                                                                                                          C                11                            =                                                C                                      1                    ⁢                    G                                                  +                                  C                  M                                                                                        Eq        .                                  ⁢                  (          18          )                    
Accordingly, the total capacitance CEVEN in the EVEN mode becomes equal to the self-capacitance C1G, as shown in Eq. (18). Using the total inductance LEVEN and the total capacitance CEVEN in Eqs. (15) and (18) above, ZEVEN and TDEVEN may be defined as:
                              Z          EVEN                =                                                            L                EVEN                                            C                EVEN                                              =                                                                      L                  11                                +                                  L                  12                                                                              C                  11                                -                                  C                  12                                                                                        Eq        .                                  ⁢                  (          19          )                                                  TD          EVEN                =                                                            L                EVEN                            ⁢                              C                EVEN                                              =                                                    (                                                      L                    11                                    +                                      L                    12                                                  )                            ⁢                              (                                                      C                    11                                    -                                      C                    12                                                  )                                                                        Eq        .                                  ⁢                  (          20          )                    
As mentioned above, the characteristic impedance of transmission line varies depending on adjacent transmission lines and signal modes (signal transmission modes) due to the influence of coupling between the adjacent transmission lines, which gives rise to a difference in the transmission rates of signals. In result, such difference in the transmission rates acts as a factor that impairs a timing margin.
FIG. 2 is a view showing several types of crosstalk generated depending on signal transmission modes.
Referring to FIG. 2, signal modes between adjacent transmission lines can be largely classified into an EVEN mode 310 and an ODD mode 320. When there are two transmission lines, the EVEN mode 310 is when signals having the same phase and the same amplitude are applied to the two transmission lines, respectively. On the other hand, the ODD mode 320 is when signals having a 180-degree phase difference but the same amplitude are applied to the two transmission lines, respectively. In addition, when there is a variation in only a signal on any one of the two transmission lines, it is regarded that the EVEN mode 310 and the ODD mode 320 overlap each other, which is called an overlap mode 300.
Now, what effect signal modes between two transmission lines would have on transmission signals on those transmission lines will be described with reference to FIG. 2.
First, referring to an EVEN mode model 311, when the EVEN mode is set between two transmission lines, it can be seen, from the signal shape 312 of the EVEN mode, the fact that the transmission rate of signal is the slowest. Further, referring to an ODD mode model 321, when the ODD mode is set between two transmission lines, it can be found, from the signal shape 322 of the ODD mode, the fact that the transmission rate of signal is the fastest. Lastly, referring to an overlap mode model 301, when the ODD mode is built up between two transmission lines, it can be found, from the signal shape 302 of the overlap mode, the fact that the signal shapes 312 and 322 of the EVEN mode and the ODD mode overlap each other.
As discussed above, there occurs a difference in the transmission rates of signals, i.e., skew, depending on signal modes between adjacent transmission lines. Among various methods developed to compensate such skew, the most frequently used method is a Delayed Locked Loop (DLL). However, this type of devices compensates such skew by synchronizing with a reference signal such as a clock, without having to consider which type of crosstalk occurred at all. Therefore, such devices require numerous control signals therein, thus making their circuits complicated and wider in area, together with much power consumption.